c Compute de Broglie wavelength for neutron energies i E 0025 MeV ii E 1 MeV
Question
Basic Answer
Step 1: Define the de Broglie wavelength formula
The de Broglie wavelength (λ) is given by the formula:
λ = h / p
where:
- h is Planck’s constant (6.626 x 10^-34 Js)
- p is the momentum of the particle
For a particle with mass m and kinetic energy E, the momentum can be expressed as:
p = √(2mE)
Step 2: Convert energy to Joules
First, we need to convert the given energies from MeV to Joules. 1 MeV = 1.602 x 10^-13 J.
(i) E = 0.025 MeV = 0.025 * 1.602 x 10^-13 J = 4.005 x 10^-15 J
(ii) E = 1 MeV = 1 * 1.602 x 10^-13 J = 1.602 x 10^-13 J
Step 3: Calculate the momentum for each energy
The mass of a neutron (m) is approximately 1.675 x 10^-27 kg. We can now calculate the momentum for each case using p = √(2mE):
(i) p₁ = √(2 * 1.675 x 10^-27 kg * 4.005 x 10^-15 J) ≈ 3.66 x 10^-21 kg m/s
(ii) p₂ = √(2 * 1.675 x 10^-27 kg * 1.602 x 10^-13 J) ≈ 7.31 x 10^-20 kg m/s
Step 4: Calculate the de Broglie wavelength for each energy
Now we can calculate the de Broglie wavelength using λ = h / p:
(i) λ₁ = (6.626 x 10^-34 Js) / (3.66 x 10^-21 kg m/s) ≈ 1.81 x 10^-13 m
(ii) λ₂ = (6.626 x 10^-34 Js) / (7.31 x 10^-20 kg m/s) ≈ 9.06 x 10^-15 m
Final Answer
(i) For E = 0.025 MeV, the de Broglie wavelength is approximately 1.81 x 10^-13 m.
(ii) For E = 1 MeV, the de Broglie wavelength is approximately 9.06 x 10^-15 m.
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